[1] Define F=LN*MK (* denoting intersection), J=ML*KN. Line FJ is the polar of E (see the basic figure for polar construction in Polar2.html ).

[2] By the same reference FE is the polar of J and JE is the polar of F.

[3] Line ML is the polar of A and passes through J. By the reciprocity of polars, the polar of J which is EF will pass through A. Analogously it will pass through C. This shows the concurence of the four lines {MN,KL,AC,BD} at E (this is a special case of Brianchon's theorem, see Brianchon2.html ).

[4] By the definition of polarity points (K,N,J,J') make a harmonic division on line NK which is projected from C to (G,H,J,F). Hence these points make also a harmonic division.

[5] Reading the figure from the viewpoint of the inscribed quadrangle LMKN we obtain the specialization of Pascal's theorem for inscriptible quadrilaterals: The intersection points {G,H} of opposite sides and {F,J} of tangents at opposite vertices are on a line and make on it a harmonic division.

[2] There are inverses to the theorems of Pascal and Brianchon for quadrangles inscribed/circumscribed in conics. The inverse of Pascal's theorem for quadrangles is shown in Pascal2.html . The assumptions for this inverse is that the four points {M,K,L,N} are such that on line FJ there are given two points {G,H} which are harmonic conjugate to {F,J}. Then there is a conic circumscribing MKNL and having opposite tangents intersecting at those {G,H}. Obviously the conic is a circle when {G,H} coincide with the intersections of FJ with the medial lines of {KL,MN} and in addition these points are harmonic conjugate to {F,J}. Anyway there are simpler conditions to test the cyclicity of MKNL, but it is interesting to see all conics circumscribing the quadrangle. They are parameterized by the position of point G on line FJ (since H is then determined from G). The figure below shows such a family of conics circumscribing the quadrangle MKNL.

For all these conics the pole of line FJ is the same point E=KL*MN.

[3] Given the position of G below the construction of the concrete conic is done by an appropriate member of the (bitangent) family of conics generated by lines {CM,CL} considered as a degenerate conic and the line ML (considered as a double line degenerated conic). The member is the one passing through one of the other two points {K,N}.

[2] The duality is between the points of the line FJ and the bundle of lines (often denoted by E*) at E which is the pole of FJ. To every pair of conjugate points {G,H} to {F,J} corresponds a circumconic and to every pair of conjugate lines {EL,EN} to {EF,EJ} corresponds an inconic.

[3] The above figure illustrates the case of all inconics to quadrilateral ABCD by determining each one of them through two points {M',L'} harmonic conjugate with respect to {F,J}. These points define the required harmonic bundle at E, E(D,C,N,L) by which the contact points {L,N,K,M} are determined.

[4] The concrete conic in the above figure is constructed is defined as an appropriate member of the (bitangent) family of conics generated by lines {CM,CL} considered as a degenerate conic and the line ML (considered as a double line degenerated conic). The member is the one passing through one of the other two points {K,N}.

[2] Analogous is the discussion in [3] concerning this time the conics inscribed in a given quadrilateral ABCD. From this discussion follows also easily the inverse of Brianchon's theorem for quadrilaterals inscriptible in circles/conics:

Given the quadrilateral ABCD, let E be the intersection point of its diagonals and {G,H} the intersection points of pairs of opposite sides. Let also {F,J} be the intersection points of the diagonals with line GH. Then for every pair (EL,EN) of lines through E, conjugate with respect to (EF,EG) there is a conic tangent to ABCD and having lines {EL,EN} as lines of contacts of the quadrilateral with the conic.

[3] Section [3] gives also a structural view to the problem of finding a conic tangent to four given lines and passing through a point lying on one of them (L say). The given point determines simply the pair of lines (EL,EN) conjugate to (EF,EG) and needed for the definition of the conic.

[4] In the file ParabolaInscribedQuad.html I discuss the problem of finding the parabola inscribed in the quadrilateral.

CircumscriptibleQuadrilateral.html

ParabolaInscribedQuad.html

Pascal2.html

Polar2.html

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